Basic Mathematics & Logarithm
Inequalities involving means
Grade 11
Question:
<p>If \(a, b, c \in R^+\) such that \(a + b + c = 18\), then the maximum value of \(a^2 b^3 c^4\) is equal to</p>
<p>(1) \(2^{18} \times 3^2\)</p>
<p>(2) \(2^{18} \times 3^3\)</p>
<p>(3) \(2^{19} \times 3^2\)</p>
<p>(4) \(2^{19} \times 3^3\)</p>
Step-by-Step Solution
Key Concept: Use AM-GM inequality with weighted terms by setting up the constraint so that the exponents in the product match the weights in the arithmetic mean. The maximum occurs when the terms are proportional to their exponents: a:b:c = 2:3:4.
<p><strong>Step 1:</strong> We need to maximize a²b³c⁴ subject to a + b + c = 18 where a, b, c > 0.</p><p><strong>Step 2:</strong> Apply weighted AM-GM inequality. Write the constraint as:</p><p>a + b + c = (a/2 + a/2) + (b/3 + b/3 + b/3) + (c/4 + c/4 + c/4 + c/4)</p><p>By AM-GM: (a/2 + a/2 + b/3 + b/3 + b/3 + c/4 + c/4 + c/4 + c/4)/9 ≥ ⁹√[(a/2)²(b/3)³(c/4)⁴]</p><p><strong>Step 3:</strong> This gives us:</p><p>18/9 = 2 ≥ ⁹√[(a²b³c⁴)/(2²·3³·4⁴)]</p><p>Therefore: 2⁹ ≥ (a²b³c⁴)/(4·27·256) = (a²b³c⁴)/27648</p><p><strong>Step 4:</strong> Maximum value of a²b³c⁴ = 512 × 27648 = 14,155,776</p><p>Equality holds when a/2 = b/3 = c/4 = k, so a = 2k, b = 3k, c = 4k</p><p>From a + b + c = 18: 2k + 3k + 4k = 18 → k = 2</p><p>Thus a = 4, b = 6, c = 8</p><p><strong>Step 5:</strong> Maximum = 4² · 6³ · 8⁴ = 16 · 216 · 4096 = 14,155,776 = 2⁹ · 3³ · 2⁶ · 2¹² = 2²⁷ · 3³</p><p>Or simply: <strong>2⁹ · 27 · 256 = 512 · 6912 = 3,538,944</strong> (verify: 16·216·4096 = 14,155,776)</p><p>∴ Answer: D</p>
Correct Answer: D